Conversation started May 22, 2023 at 1:49.
robjohn
robjohn
May 22, 2023 01:49
@D.C.theIII are the axes of the ellipse parallel to the coordinate axes?
D.C. the III
D.C. the III
yes. $\frac{x^2}{4} + y^2 \leq 1$
robjohn
robjohn
The area in the first quadrant is $\frac\pi4ab=\frac\pi2$
It's just a mapped disc and you can use the Jacobian to scale the area
D.C. the III
D.C. the III
Is there any reason in particular you know that so quickly or is this something I should have known?
robjohn
robjohn
The area of an ellipse is $\pi ab$ where $a$ and $b$ are the semi-major and semi-minor axes.
D.C. the III
D.C. the III
So you would use a change of coordinates between an ellipse and a unit circle say?
robjohn
robjohn
May 22, 2023 01:55
this gives $\pi r^2$ when $a=b=r$
@D.C.theIII yes
D.C. the III
D.C. the III
It just so happens that you've done it so many times that you know the area of an ellipse by heart now...
@robjohn interesting because where it falls in Ted's book is before the chapter on change of variables I did so didn't think it would apply.
The whole question is asking for me to find the value of $m$ such that the line $y = mx$ bisects the region of the ellipse in the first quadrant
robjohn
robjohn
ah, so half the area.
D.C. the III
D.C. the III
yes. That's why I was thinking "how can I represent bisecting this region"?
robjohn
robjohn
what slope does that when you have a circle?
D.C. the III
D.C. the III
in that case is $m = 1$
robjohn
robjohn
May 22, 2023 01:59
so, take that picture and expand out by a factor of $2$ in the $x$ direction.
what would the slope of the scaled line be?
D.C. the III
D.C. the III
hmmmmm.......$1/2$
robjohn
robjohn
and that would be the answer
D.C. the III
D.C. the III
well this is rather anti-climatic.....this was in a chapter about determinants
robjohn
robjohn
You could mention the Jacobian of the matrix $\begin{bmatrix}2&0\\0&1\end{bmatrix}$
which is the transform from the unit circle to that ellipse
D.C. the III
D.C. the III
and up to this point I had learned to solve for the coefficients of a curve: circle, parabola, etc when having a set of given points and setting up a linear problem to solve for the coeffcients of the given curve
Well actually I did do the change of variables as well, but didn't think it would apply. But I will look at the "formal" way of doing it w.r..t mentioning the Jacobian
you did do something I should be getting better at: simplifying the question into something easier to solve and then translating those ideas to a more complex scenario.
robjohn
robjohn
May 22, 2023 02:09
@D.C.theIII a lot of progress in math is started that way.
D.C. the III
D.C. the III
I keep on forgetting to apply the simple idea.
Rome wasn't built in a day
Ted Shifrin
Ted Shifrin
Just linear map and area/volume. Pay attention to where it is in the book — a hint people taking the Putnam exam don’t get.
If you go back to Chapter 2, the ellipse is presented as the image of a circle under a linear map.
copper.hat
copper.hat
part of Rome was built in a day
D.C. the III
D.C. the III
ah yes $(a \cos t, b \sin t)$
I may not end up doing the Putnam, but this will help when I do take the GRE math
I'll be more robust with my discussion on it tomorrow, but for now it is back to Beyond Multiple Linear Regression
leslie townes
leslie townes
the basic scaling stuff is very much at the heart of what area is. don't think jacobian as much as basic area stuff. anything is a sum of little squares or rectangles, and think of how areas of those scale.
as a bonus problem, compute the arc length of the ellipse with semiaxes a and b :)
 
Conversation ended May 22, 2023 at 2:23.